Induction 1^2+3^2+5^2+....+ (2n-1)^2=n/3(4n^2-1)

[markosheehan]

prove by induction that 1^2+3^2+5^2+....+ (2n-1)^2=n/3(4n^2-1) when n is a natural number. i cant work this out i tried leting k/3(4k^2-1)+(2k-1)^2 and simplyfying but i got nohere

[Guest]

You almost had it (the [tex](2k-1)^2[/tex] term should be [tex](2k+1)^2[/tex]).

It is easy to check the formula holds for [tex]n=1[/tex].
Assume it holds for [tex]n=k[/tex],
this means that [tex]1^2+\ldots+(2k-1)^2 = \frac{k(4k^2-1)}{3}[/tex],
so [tex]1^2+\ldots+(2k-1)^2+(2k+1)^2 = \frac{k(4k^2-1)}{3} +(2k+1)^2[/tex].
[tex]=\frac{4k^3+12k^2+11k+3}{3}[/tex]
[tex]=\frac{(k+1)(4(k+1)^2-1)}{3}[/tex]
So the formula holds for [tex]n=k+1[/tex], and by induction all [tex]n[/tex].

Hope this helped,

R. Baber.